What is the value of arctan(cos(0)) ?

The value of arctan(cos(0)) is pi/4

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arctan(cos(0))

arctan (cos (0))

\frac{π}{4}

Decimal Notation

45

Solution steps

arctan (cos (0))

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= arctan (1)

= arctan (1)

Use the following trivial identity:

arctan (1) = \frac{π}{4}

arctan (1)

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4}

= \frac{π}{4}

\frac{5}{tan ( 3 0 ^{\circ} )}

cos (1 5^{\circ}) cos (7 5^{\circ}) - sin (1 5^{\circ}) sin (7 5^{\circ})

50 \cdot sin (3 0^{\circ})

\frac{1}{2} sin (2 π)

- sin (- \frac{π}{4})